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A solution of \(\mathrm{Na}_{2} \mathrm{SO}_{4}\) is added dropwise to a solution that is \(0.010 \mathrm{M}\) in \(\mathrm{Ba}^{2+}\) and \(0.010 \mathrm{M}\) in \(\mathrm{Sr}^{2+}\). (a) What concentration of \(\mathrm{SO}_{4}{ }^{2-}\) is necessary to begin precipita- tion? (Neglect volume changes. \(\mathrm{BaSO}_{4}: K_{s p}=1.1 \times 10^{-10}\) \(\mathrm{SrSO}_{4}: K_{s p}=3.2 \times 10^{-7} .\) (b) Which cation precipi- tates first? (c) What is the concentration of \(\mathrm{SO}_{4}^{2-}\) when the second cation begins to precipitate?

Short Answer

Expert verified
Ba虏鈦 ions precipitate first, and the concentration of SO鈧劼测伝 ions when the second cation, Sr虏鈦, begins to precipitate is \(3.2 \times 10^{-5} M\).

Step by step solution

01

Write the solubility product expressions for each precipitate.

For the given compounds, the solubility product expressions are: For BaSO鈧: \(K_{sp}=[\mathrm{Ba}^{2+}][\mathrm{SO}_{4}^{2-}] = 1.1 \times 10^{-10}\) For SrSO鈧: \(K_{sp}=[\mathrm{Sr}^{2+}][\mathrm{SO}_{4}^{2-}] = 3.2 \times 10^{-7}\)
02

Determine the concentration of SO鈧劼测伝 required to precipitate each cation.

We will use the Ksp expressions for both precipitates and the given concentrations of Ba虏鈦 and Sr虏鈦 ions. For BaSO鈧: \([SO_{4}^{2-}] = \cfrac{K_{sp}}{[Ba^{2+}]} = \cfrac{1.1 \times 10^{-10}}{0.010} = 1.1 \times 10^{-8} \mathrm{M}\) For SrSO鈧: \([SO_{4}^{2-}] = \cfrac{K_{sp}}{[Sr^{2+}]} = \cfrac{3.2 \times 10^{-7}}{0.010} = 3.2 \times 10^{-5} \mathrm{M}\)
03

Determine which cation will precipitate first.

Comparing the concentrations of SO鈧劼测伝 required for precipitation: BaSO鈧: \(1.1 \times 10^{-8} \mathrm{M}\) SrSO鈧: \(3.2 \times 10^{-5} \mathrm{M}\) Since the required concentration of SO鈧劼测伝 is lower for BaSO鈧, Ba虏鈦 will precipitate first.
04

Calculate the concentration of SO鈧劼测伝 when the second cation begins to precipitate.

Since Ba虏鈦 precipitates first, we know that the concentration of SO鈧劼测伝 at this point is \(1.1 \times 10^{-8} \mathrm{M}\). In order for Sr虏鈦 to precipitate, the concentration of SO鈧劼测伝 must reach \(3.2 \times 10^{-5} \mathrm{M}\). So, the concentration of SO鈧劼测伝 when the second cation begins to precipitate is \(3.2 \times 10^{-5} M\). In conclusion, Ba虏鈦 ions precipitate first, and the concentration of SO鈧劼测伝 ions when the second cation, Sr虏鈦, begins to precipitate is \(3.2 \times 10^{-5} M\).

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Precipitation Reactions
Precipitation reactions occur when two soluble ionic compounds in a solution form an insoluble solid called a precipitate. These reactions are guided by the concept of solubility product constant, or \( K_{sp} \), which helps determine when a compound will begin to precipitate from solution.
In these reactions, ions from each reactant are mixed in a solution and if the resulting product exceeds the solubility product constant, a precipitate forms. The concentration of ions and the specific \( K_{sp} \) values for the compounds involved govern when precipitation occurs.
A practical example is the gradual addition of a sodium sulfate (Na鈧係O鈧) solution to a mixture of barium ions \((\text{Ba}^{2+})\) and strontium ions \((\text{Sr}^{2+})\). Understanding which ion will precipitate first helps predict the sequence and properties of the formed precipitates.
Barium Sulfate
Barium sulfate (BaSO鈧) is a compound known for its low solubility in water, which is reflected in its solubility product constant, \( K_{sp} = 1.1 \times 10^{-10} \). This low \( K_{sp} \) value indicates that BaSO鈧 will precipitate out of a solution at a relatively low concentration of sulfate ions \((\text{SO}_{4}^{2-})\).
In the context of precipitation reactions, Ba虏鈦 ions will combine with \(\text{SO}_{4}^{2-}\) ions as soon as their product exceeds the \( K_{sp} \) value. For example, when \([\text{SO}_{4}^{2-}]\) reaches \(1.1 \times 10^{-8} \text{ M}\), BaSO鈧 starts precipitating out of solution.
This makes barium sulfate an ideal compound for studying the sequence of precipitation reactions, as its early precipitation can be easily predicted and observed.
Strontium Sulfate
Strontium sulfate (SrSO鈧) has a relatively higher solubility than barium sulfate, with a \( K_{sp} = 3.2 \times 10^{-7} \). This means that it will precipitate from a solution only when the concentration of \( \text{SO}_{4}^{2-} \) reaches a higher threshold of \( 3.2 \times 10^{-5} \text{ M} \).
In mixtures containing both Ba虏鈦 and Sr虏鈦 ions, strontium sulfate precipitates after barium sulfate, due to its higher \( K_{sp} \). The delayed precipitation allows for the sequential removal of ions based on their solubility product constants.
This higher solubility can be leveraged in chemical processes to control the order of precipitation, making strontium sulfate particularly useful for studying the dynamics of ionic solubility and compound formation.
Cation Precipitation Sequence
Cation precipitation sequence refers to the order in which different metal ions precipitate from a solution. The sequence is determined primarily by the solubility products of the compounds they form with the anions present in the solution.
In cases where solutions contain multiple cations that could form precipitates with a common anion, like \( \text{SO}_{4}^{2-} \), the cation with the lower \( K_{sp} \) value will precipitate first. For the scenario with barium and strontium ions, Ba虏鈦 precipitates first because BaSO鈧 has a lower \( K_{sp} \) than SrSO鈧.
Understanding this sequence is crucial in industrial processes and laboratory settings, as it aids in the selective separation and purification of specific ions based on their chemical properties.

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Most popular questions from this chapter

A concentration of \(10-100\) parts per billion (by mass) of \(\mathrm{Ag}^{+}\) is an effective disinfectant in swimming pools. However, if the concentration exceeds this range, the \(\mathrm{Ag}^{+}\) can cause adverse health effects. One way to main-tain an appropriate concentration of \(\mathrm{Ag}^{+}\) is to add a slightly soluble salt to the pool. Using \(K_{s p}\) values from Appendix \(\mathrm{D}\), calculate the equilibrium concentration of \(\mathrm{Ag}^{+}\) in parts per billion that would exist in equilibrium with (a) \(\mathrm{AgCl}\), (b) \(\mathrm{AgBr}\), (c) AgI.

A buffer solution contains \(0.10 \mathrm{~mol}\) of acetic acid and $0.13 \mathrm{~mol}\( of sodium acetate in \)1.00 \mathrm{~L}$. (a) What is the \(\mathrm{pH}\) of this buffer? (b) What is the \(\mathrm{pH}\) of the buffer after the addition of \(0.02\) mol of KOH? (c) What is the pH of the buffer after the addition of \(0.02 \mathrm{~mol}\) of \(\mathrm{HNO}_{3}\) ?

Describe the solubility of \(\mathrm{CaCO}_{3}\) in each of the following solutions compared to its solubility in water: (a) in \(0.10 \mathrm{M} \mathrm{NaCl}\) solution; \((\mathrm{b})\) in \(0.10 \mathrm{M} \mathrm{Ca}\left(\mathrm{NO}_{3}\right)_{2}\) solution; (c) \(0.10 \mathrm{M} \mathrm{Na}_{2} \mathrm{CO}_{3}\); (d) \(0.10 \mathrm{M}\) HCl solution. (Answer same, less soluble, or more soluble.)

A buffer is prepared by adding \(7.00 \mathrm{~g}\) of ammonia \(\left(\mathrm{NH}_{3}\right)\) and \(20.0 \mathrm{~g}\) of ammonium chloride \(\left(\mathrm{NH}_{4} \mathrm{Cl}\right)\) to enough water to form \(2.50 \mathrm{~L}\) of solution. (a) What is the \(\mathrm{pH}\) of this buffer? (b) Write the complete ionic equation for the reaction that occurs when a few drops of nitric acid are added to the buffer. (c) Write the complete ionic equation for the reaction that occurs when a few drops of potassium hydroxide solution are added to the buffer.

The value of \(K_{s p}\) for \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) is \(2.1 \times 10^{-20}\). The \(\mathrm{AsO}_{4}{ }^{3-}\) ion is derived from the weak acid \(\mathrm{H}_{3} \mathrm{AsO}_{4}\) \(\left(\mathrm{pK}_{a 1}=2.22 ; \mathrm{pK}_{a 2}=6.98 ; \mathrm{pK}_{a 3}=11.50\right) .\) When asked to calculate the molar solubility of \(\mathrm{Mg}_{3}\left(\mathrm{AsO}_{4}\right)_{2}\) in water, a student used the \(K_{s p}\) expression and assumed that \(\left[\mathrm{Mg}^{2+}\right]=1.5\left[\mathrm{AsO}_{4}^{3-}\right]\). Why was this a mistake?

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